Problem on MM equation
How to obtain relation between Vm and Km,given k(sec^-1) = Vmax/mg of enzyme x molecular weight x 1min/60 sec S* = 4.576(log K -10.753-logT+Ea/4.576T).
Expert
According to M.M Equation, V = Vmax x {S]/Km + {S] Since Vmax = ksec^-1 {Et}V = Ksec^-1{Et} x {S}/Km + {S}Substituting the given value of Ksec^-1 in the above equation,we get V = Vmax x mol.wt x 1min/60 sec x {S}/mg of enzyme / Km +{S]V = Vmax x mol.wt x {S}/mg of enzyme/Km +{S}Substituting the given value of {S}V = Vmax x mol.wt x 4.576(log K -10.753-logT+Ea/4.576T)/mg of enzyme/ Km + 4.576(log K -10.753-logT+Ea/4.576T)
Choose the right answer from following. When 6gm urea dissolve in180gm H2O . The mole fraction of urea is : (a)10/ 10.1 (b)10.1/10 (c)10.1/ 0.1 (d) 0.1/ 10.1
Write a short note to describe how to test a gas to see if it was hydrogen or not?
How much of NaOH is needed to neutralise 1500 cm3 of 0.1N HCl (given = At. wt. of Na =23): (i) 4 g (ii) 6 g (iii) 40 g (iv) 60 g
Choose the right answer from following. The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be: (a) 18.0 (b) 342 (c) 60 (d) 180
Illustrate the 3 facts on evaporation?
At low temperatures, mixtures of water and methane can form a hydrate (i.e. a solid containing trapped methane). Hydrates are potentially a very large source of underground trapped methane in the pole regions but are a nuisance when they form in pipelines and block th
The equation S = k in W relates entropy to W, a measure of the number of different molecular level arrangements of the system.In the preceding developments it was unnecessary to attempt to reach any "explana
brief discription of relative lowering of vapour pressure
In some two component, solid liquid systems, a eutectic mixture forms.Consider, now a two component system at some fixed pressure, where the temperature range treated is such as to include formation of one or more solid phases. A simple behavior is shown b
The free energy of a component of a liquid solution is equal to its free energy in the equilibrium vapour.Partial molal free energies let us deal with the free energy of the components of a solution. We use these free energies, or simpler concentration ter
18,76,764
1932125 Asked
3,689
Active Tutors
1429484
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!